(a) The number of vacancies per cubic centimeter is 1.157 X 10²⁰
(b) ρ = n X (AM) / v X Nₐ
<u>Explanation:</u>
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Given-
Lattice parameter of Li = 3.5089 X 10⁻⁸ cm
1 vacancy per 200 unit cells
Vacancy per cell = 1/200
(a)
Number of vacancies per cubic cm = ?
Vacancies/cm³ = vacancy per cell / (lattice parameter)³
Vacancies/cm³ = 1 / 200 X (3.5089 X 10⁻⁸cm)³
Vacancies/cm³ = 1.157 X 10²⁰
Therefore, the number of vacancies per cubic centimeter is 1.157 X 10²⁰
(b)
Density is represented by ρ
ρ = n X (AM) / v X Nₐ
where,
Nₐ = Avogadro number
AM = atomic mass
n = number of atoms
v = volume of unit cell
Answer:
a) 70.29 %
b) 37%
Explanation:
percent reduction can be found from:
PR = 100*(π(do/2)^2-π(df/2)^2)/π(do/2)^2
= 100*(π(11.34/2)^2-π(6.21/2)^2)/π(11.34/2)^2
=70.29 %
percent elongation can be found from:
EL =L_f - Lo/Lo*100
= (73.17 -53.3/53.3)*100
= 37%
Answer:
the hurts my brain sorry bud cant help
Explanation:
Answer:
Observational Skills
Explanation:
Observing the area also known as scanning the scene
Answer:
vB = - 0.176 m/s (↓-)
Explanation:
Given
(AB) = 0.75 m
(AB)' = 0.2 m/s
vA = 0.6 m/s
θ = 35°
vB = ?
We use the formulas
Sin θ = Sin 35° = (OA)/(AB) ⇒ (OA) = Sin 35°*(AB)
⇒ (OA) = Sin 35°*(0.75 m) = 0.43 m
Cos θ = Cos 35° = (OB)/(AB) ⇒ (OB) = Cos 35°*(AB)
⇒ (OB) = Cos 35°*(0.75 m) = 0.614 m
We apply Pythagoras' theorem as follows
(AB)² = (OA)² + (OB)²
We derive the equation
2*(AB)*(AB)' = 2*(OA)*vA + 2*(OB)*vB
⇒ (AB)*(AB)' = (OA)*vA + (OB)*vB
⇒ vB = ((AB)*(AB)' - (OA)*vA) / (OB)
then we have
⇒ vB = ((0.75 m)*(0.2 m/s) - (0.43 m)*(0.6 m/s) / (0.614 m)
⇒ vB = - 0.176 m/s (↓-)
The pic can show the question.
